Transcript Recombinants and Linkage Maps

Many fruit fly genes were mapped initially using recombination frequencies
T.H. Morgan did experiments with fruit flies to see how linkage affects
the inheritance of two different characters
~ Linked genes that are close together on the same chromosome
do not assort independently
~ Genes that assort independently are either:
- on separate chromosomes OR
- are far apart on the same chromosome
Y
II
I
X
Short
aristae
0
Long aristae
(appendages
on head)
IV
III
Mutant phenotypes
Black
Cinnabar Vestigial Brown
body
eyes
wings
eyes
48.5 57.5 67.0
Gray
Normal
Red
body
wings
eyes
Wild-type phenotypes
104.5
Red
eyes
Copyright © 2002 Pearson Education, Inc., publishing as Benjamin Cummings
Linkage Mapping: Using Recombination Data
•Cross true breeding parents of different phenotypes
•Cross heterozygous F1
organisms with
pure-breeding recessives
(like a TEST CROSS)
• Count recombinants
(ones that look different
from parental phenotype)
• Geneticists say that Linked genes
exhibit a recombination frequency
less than 50%.
• A linkage map
– Is the actual map of a chromosome based on
recombination frequencies
APPLICATION
A linkage map shows the relative locations of genes along a chromosome.
TECHNIQUE
A linkage map is based on the assumption that the probability of a crossover between two
genetic loci is proportional to the distance separating the loci. The recombination frequencies used to construct
a linkage map for a particular chromosome are obtained from experimental crosses, such as the cross depicted
in Figure 15.6. The distances between genes are expressed as map units (centimorgans), with one map unit
equivalent to a 1% recombination frequency. Genes are arranged on the chromosome in the order that best fits the data.
RESULTS In this example, the observed recombination frequencies between three Drosophila gene pairs
(b–cn 9%, cn–vg 9.5%, and b–vg 17%) best fit a linear order in which cn is positioned about halfway between
the other two genes:
Recombination
frequencies
9.5%
9%
17%
Chromosome b
Figure 15.7
cn
vg
The b–vg recombination frequency is slightly less than the sum of the b–cn and cn–vg frequencies because double
crossovers are fairly likely to occur between b and vg in matings tracking these two genes. A second crossover
would “cancel out” the first and thus reduce the observed b–vg recombination frequency.
Essential knowledge 3.A.3.b
Evidence of student learning is a demonstrated understanding of each of the following:
3. The pattern of inheritance (monohybrid, dihybrid, sex-linked, and genes linked on the
same homologous chromosome) can often be predicted from data that gives the parent
genotype/ phenotype and/or the offspring phenotypes/genotypes.
female flies with red eyes (wild type)
male flies with white eyes
The F1 generation all had red eyes
F2 generation showed the 3:1 red:white eye ratio,
but only males had white eyes
T.H. Morgan proposed that the white eye mutation
was carried on X chromosome
A Wild type fruit fly (heterozygous for gray body
and normal wings) is mated with a black fly with
vestigial wings.
OFFSPRING:
778- wild type
785- black-vestigial
158- black- normal wings
162- gray body-vestigial wings
What is the recombination frequency between
these genes?
A Wild type fruit fly (heterozygous for gray body
and red eyes) is mated with a black fly with
purple eyes.
OFFSPRING:
721- gray body/red eyes
751- black body/purple eyes
49- gray body/purple eyes
45- black body/red-eyes
What is the recombination frequency between
these genes?
Determine the sequence of genes along a
chromosome based on the following
recombination frequencies:
A-B = 8 map units
A-C = 28 map units
A-D = 25 map units
B-C = 20 map units
B-D = 33 map units
Determine the sequence of genes along a
chromosome based on the following
recombination frequencies:
A-C = 20 map units
A-D = 10 map units
B-C = 15 map units
B-D = 5 map units